Gauss' Lemma is stated and proved on pages 303-304 of Dummit and Foote (see attachment)

================================================== =========================================

Gauss Lemma

Let R be a UFD with field of fractions F and let . If p(x) is reducible in F[x] then p(x) is reducible in R[x].

More precisely, if p(x) = A(x)B(x) for some nonconstant polynomials , then there are nonzero elements such that rA(x) = a(x) and sB(x) = b(x) both lie in R[x] and p(x) = a(x)b(x) is a factorization in R[x].

================================================== =========================================

In the proof of Gauss' Lemma on page 304 (see attachment) we find:

"Assume d is not a unit and write d as a product of irreducibles in R, say . Since is irreducible in R, the ideal is prime (cf Proposition 12, section 8.3 - see attachment), so by Proposition 2 above (see attachment) the ideal is prime in R[x] and is an integral domain.

Reducing the equation dp(x) = a'(x)b'(x) modulo we obtain the equation , hence one of the two factors, say must be zero. ... ... (see attachment) .. "

================================================== ===========================================

Problem!

In the proof we read:

"Since is irreducible in R, the ideal is prime (cf Proposition 12, section 8.3 - see attachment), so by Proposition 2 above (see attachment) the ideal is prime in R[x] and is an integral domain. ..."

I can see that this is the case, BUT why is this needed and how does the rest of the proof connect to this???

The next part of the proof is

"Reducing the equation dp(x) = a'(x)b'(x) modulo we obtain the equation , hence one of the two factors, say must be zero. ... ... (see attachment) .. "

But to do this we only need to divide by and take the remainder as defining the coset. Is the section above confirming in some way that we can divide successfully???

Can someone please clarify?

Peter